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13    EXTENSIONS OF HERMITIAN OPERATORS       44!

and is an eigenvalue of A. The essential spectrum of A is therefore the set of non-
isolated points of SpO*). (Reduce to the case A = 0, and by using Problem 1 of Section
15.12 reduce further to the case where Kvr(A) = {0}; then show that 0 is a regular
value for A, and use (15.12.11).)

(c) If H is a closed hermitian operator of defect (m, 77), then $p(H) contains the half-
plane Jz J> 0 (resp. ./z <J 0) if m > 0 (resp. /; > 0). The essential spectrum of H is
contained in R. If m and n are finite, and if HI is a closed hermitian operator which
extends H, then the essential spectra of H and HI are the same (Section 15.12,
Problem l(h)).

12.    Let H be an unbounded hermitian operator on a separable Hilbert space E.

(a)    If Hj, is the restriction of H* to the subspace dom(Ht) = dom(H) -f Ker(#*),
show that JEfi is hermitian.

(b)    Show that if lm(H) is closed in E, then H is self-adjoint. (If xedomttff),
show that H*  x is orthogonal to KerCfiT *), and therefore H* - x e lm(H) (Section
15.12, Problem l(c)); deduce that x e dom(HM If His closed and if there exists a
real number which does not belong to the essential spectrum of H, deduce that the
defects of H are equal.

(c)    Show that Efit (resp. Et) (notation of (15.13.6) is the intersection of Ej (resp. EH)
with the orthogonal supplement of Ker(#*). Deduce that if there exists A e R not
belonging to the essential spectrum of H, and if the defects of H (which are necessarily
equal) have a finite value m, then dim(Ker(jfiT*  A/)) J> m.

(d)    Suppose that H is closed, and let x be an eigenvector of H* corresponding to a
real eigenvalue A. Put x = XQ-\- y-\- z, where XQ e domCfiT), y e E$ and z 6 EH . Show
that II^H = ||z|| (reduce to the case A = 0).

(e)    Suppose that H is closed, and let A be a real number which does not belong to
the essential spectrum of H; suppose that the defects of H are equal and finite, say m,
and that dim(K&r(H - A/)) is finite, say k. Then dim(Ker(#* - A/)) = m + k. (We
may assume that A = 0. By considering the restriction of H to the orthogonal supple-
ment of Ker(#), reduce to the case where k = 0. Then deduce from (d) that Ker(7ir*)
cannot have dimension >m, by using the hypothesis Kerf-fl"*) n domCfiT) = {0}, and
complete the proof by using (c).)

13.    Let //"be a closed hermitian operator whose defects are equal and finite, say m.

(a)    If Vis defined as in (15,13.4), then the self-adjoint extensions A of H are of
the form A = /(/ + U)(I C/)"1, where U is a unitary operator extending V, such
that U(Ea) = EH', dom(/4) is therefore the direct sum of domCfiT) and the subspace
(/ /)(/*) of dimension m, contained in dom(T*).

(b)    For a real number A to be an eigenvalue of a self-adjoint extension A of H, it is
necessary and sufficient that A should be an eigenvalue of H* (use Problem 12(d)).
Show that if A e R is not an eigenvalue of H, then there exists a self-adjoint extension
A of H for which A is not an eigenvalue. (Use Problem 12(d) and (e), and choose
suitably the unitary operator U of (a).)

(c)    Suppose that m > 0. For each A 6 R, show that there exists a self-adjoint exten-
sion A of H such that A e Sp(^). (Remark that, if A0 is a self-adjoint extension of H
and A  SpC40), then it follows from Problems 12(e) and ll(c) that A is an eigenvalue
of H*.)

(d)    Let A i, A2 be two self-adjoint extensions of H. lfP+ (resp. P~) is the orthogonal
projection onto Ejf (resp. E)> show that the continuous operatoroint